Answer to Question #317608 in Statistics and Probability for miekay

Question #317608

if the population consists of numbers 1,2,3,4,5 and 6,find the mean of the samples of the samples of size 3.construct a sampling distribution


1
Expert's answer
2022-03-28T13:39:21-0400

μ(1,2,3)=(1+2+3)/3=2\mu(1,2,3)=(1+2+3)/3=2

μ(1,2,4)=(1+2+4)/3=2.3\mu(1,2,4)=(1+2+4)/3=2.3

μ(1,2,5)=(1+2+5)/3=2.7\mu(1,2,5)=(1+2+5)/3=2.7

μ(1,2,6)=(1+2+6)/3=3\mu(1,2,6)=(1+2+6)/3=3

μ(2,3,4)=(2+3+4)/3=3\mu(2,3,4)=(2+3+4)/3=3

μ(2,3,5)=(2+3+5)/3=3.3\mu(2,3,5)=(2+3+5)/3=3.3

μ(2,3,6)=(2+3+6)/3=3.7\mu(2,3,6)=(2+3+6)/3=3.7

μ(3,4,5)=(3+4+5)/3=4\mu(3,4,5)=(3+4+5)/3=4

μ(3,4,6)=(3+4+6)/3=4.3\mu(3,4,6)=(3+4+6)/3=4.3

μ(4,5,6)=(4+5+6)/3=5\mu(4,5,6)=(4+5+6)/3=5

f(2)=f(2.3)=f(2.7)=f(3.3)=f(3.7)=f(4)=f(4.3)=f(5)=1/10

f(3)=2/10

E(x)=xf(x)=1/10(2+2.3+2.7+3.3+3.7+4+4.3+5)+2/10×3=2.73+0.6=3.33E(x)=\sum xf(x)=1/10(2+2.3+2.7+3.3+3.7+4+4.3+5)+2/10\times3=2.73+0.6=3.33

σ2=x2f(xf)2=1/10(4+5.29+7.29+10.89+13.69+16+18.49+25)+2/10×93.332=11.86511=0.865\sigma^2=\sum x^2f - (\sum xf)^2=1/10(4+5.29+7.29+10.89+13.69+16+18.49+25)+2/10\times9-3.33^2=11.865-11=0.865





Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog